Let $f \in \mathbb{Z}[x, y]$ be a polynomial. Suppose that the list of terms of $f$ do not involve the $y$ variable except for a single $y^2$ term with some arbitrary coefficient. When is $f$ reducible?
I've been able to find the family of solutions $f(x, y) = (ax + b)^{2n} - (cy)^2$ for $a, b, c \in \mathbb{Z}$ and $n \in \mathbb{N}$, and I was able to show that this contains the only solutions for $f$ of degree 2 (i.e. when $n = 1$). I suspect there are no other $f$ that work but I'm not sure of an approach on how to go about proving it?
Let's say we have $f(x,y) = g(x,y)h(x,y)$ and let's look at the degrees of $g$ and $h$ with respect to $y$.
In the first case, one of them has degree $2$ and the other degree $0$ wrt $y$, so $f(x,y) = (ay^2+p(x))q(x)$, but assuming that coefficient of $y^2$ is an integer, it means that $q(x)$ is a constant polynomial, so $f(x,y) = aby^2 + bp(x)$. In the other direction, we'd like $b$ to be non-zero and non-invertible, i.e. $|b| \geq 2$, and we'd get $f(x,y) = b(ay^2+p(x))$, so $f$ is reducible.
In the second case, degrees of $g$ and $h$ wrt $y$ are both $1$, so $$f(x,y) = (ay+p(x))(by+q(x)) = ab y^2 + (bp(x) + aq(x))y + p(x)q(x),$$ so it's necessary that $bp(x) + aq(x) = 0$.
Let $k = \gcd(a,b)$ and $a = ka'$, $b = kb'$, so $a'$ and $b'$ are coprime. Note that from $bp(x) + aq(x) = 0$ it follows that $b'p(x) + a'q(x) = 0$, so $a'\mid p(x)$ and $b'\mid q(x)$, i.e. $a'b'\mid p(x)q(x)$. Therefore, we can factor out $a'b'$ and if $|a'b'| \geq 2$ we are back to the previous case. So, assume $|a'b'| = 1$, i.e. $a = \pm b$. Now we have $f(x,y) = \pm (a^2y^2 + p(x)q(x))$ and $p(x)+q(x) = 0$, which combined gives us $f(x,y) = \pm (a^2y^2 - p(x)^2)$.
Note that if you don't accept factorization where one of the polynomials is constant, you can ignore the first case. You can modify the argument in the second case by assuming that $f$ is primitive polynomial.